Answer
$\dfrac{6}{s^2+9}+\dfrac{24}{s^4}$
Work Step by Step
The Laplace Transform can be written as:
$L[F(t)]=\int_{0}^{\infty} e^{-st} f(t) dt $
We are given that $f(t)=2 \sin (3t)+4t^3$
Now, $L[F(t)]=\int_{0}^{\infty} e^{-st} f(t) dt \\=\int_{0}^{\infty} e^{-st} [2 \sin (3t)+4t^3] dt\\=2 [\dfrac{3}{s^2+9}]+4 [\dfrac{3 !}{s^{3+1}}] \\=\dfrac{6}{s^2+9}+\dfrac{24}{s^4}$
